Answer:
![\sf \Bigg[ \frac{ ( - 4 \times ( - 3)) }{2} \Bigg] \: and \: \Bigg[ \frac{ - ( - 5 \times 8)}{10} + 2 \Bigg]](https://tex.z-dn.net/?f=%20%20%20%5Csf%20%5CBigg%5B%20%5Cfrac%7B%20%28%20-%204%20%5Ctimes%20%28%20-%203%29%29%20%7D%7B2%7D%20%5CBigg%5D%20%5C%3A%20and%20%20%5C%3A%20%5CBigg%5B%20%5Cfrac%7B%20-%20%28%20-%205%20%5Ctimes%208%29%7D%7B10%7D%20%2B%202%20%5CBigg%5D)
Step-by-step explanation:
Given:
To find:
Two expressions that equal 6 using the given numbers
Solution:
Expression first,
Using numbers -4, 2, -3,
aligning the above numbers as,
will out put 6.
<em>Verification,</em>
Expression second,
Using numbers 10,8,2,-5
aligning the above numbers as,
will result 6.
<em>Verification</em>
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Answer: The loser's card shows 6.
Explanation: Let's start by naming the first student A and the second student B.
Since the product of A and B are either 12, 15, or 18, let's list every single possibility, the first number being A's number and the second number being B's number.
1 12
1 15
1 18
2 6
2 9
3 4
3 5
3 6
4 3
5 3
6 2
6 3
9 2
12 1
15 1
18 1
Now, the information says that A doesn't know what B has, so we can immediately cross off all of the combinations that have the integer appearing once and once ONLY off, because if it happened once only, A would know of it straight away. Now, our sample space becomes much smaller.
1 12
1 15
1 18
2 6
2 9
3 4
3 5
3 6
6 2
6 3
Using this same logic, we know that we can cross off all of the digits that occur only once in B's column.
2 6
3 6
Now, A definitely knows what number B has because there is only one number left in B. Hence, we can conclude that the loser, B, has the integer 6.
I believe that would be 64C.
